Question

有没有一种方法可以使此ELIF语句更简单，使此代码不起作用。更改的变量是APPOINTMENT

elif (rut == '80010900-0' and agental_launch != "" and 'TAL'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'IQQ'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'ANF'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'MJS'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'QTV'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'SVE'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'PMC'  in appointment) or
        (rut == '80010900-0' and agental_launch != "" and 'CHB'  in appointment):
        df.at[idx,'REBATE'] = round(int(monto_neto)*0.35)

Answer 1

将any用于最后一个变成单线的条件：

elif rut == '80010900-0' and agental_launch != "" and any(x in appointment for x in ['TAL','IQQ','rest of strings to match...']):

Answer 2

您可以这样做：

elif rut == '80010900-0' and agental_launch != "" and any(elem in appointment for elem in ['TAL','IQQ','ANF','MJS','QTV','SVE','PMC','CHB']):
     df.at[idx,'REBATE'] = round(int(monto_neto)*0.35)

如果陈述简单

2 个答案: